Good afternoon,

I am really stuck on a particular partial fraction decomposition problem.

I have:

Which easily becomes:

Now, should this be setup like?:

If this should be setup like this, I am confused about where to turn next.

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- July 12th 2010, 08:38 AMMechEngPartial Fraction Decomposition
Good afternoon,

I am really stuck on a particular partial fraction decomposition problem.

I have:

Which easily becomes:

Now, should this be setup like?:

If this should be setup like this, I am confused about where to turn next. - July 12th 2010, 09:15 AMAckbeet
For any partial fractions setup, the next thing you need to do is set your setup equal to the original fraction, get common denominators, equate like powers of x, and solve for your constants. Make sense?

- July 12th 2010, 10:02 AMMechEng
So, should that look like:

Subsequently I get:

Should I then set x=0 to get:

1 = 2B

Substituting B=1/2 back in, I get

Where do I go from here? - July 12th 2010, 10:06 AMAckbeet
Everything looks fine so far. I wouldn't actually have thought of setting x=0, but that works well in quickly obtaining one of the constants. You need to equate the coefficients of like powers of x. For the powers, for example, you have . You'll get a system of equations you'll have to solve simultaneously. Can you finish from here?

- July 12th 2010, 10:13 AMMechEng
[QUOTE=Ackbeet;536165]You need to equate the coefficients of like powers of x. [QUOTE]

I'm not sure I follow you here. My book doesn't do a particularly great job of explaining this. - July 12th 2010, 10:28 AMAckbeet
Ok, let me re-write your equation there: you have

The LHS does not need to be altered, because it already looks the way I want. The RHS I alter such that you get:

Did you follow that?

Now, in order for this equation to hold, you have to equate the coefficients of like powers of x, or else one term will start growing too quickly. That is, you can read off the following four equations immediately:

Did you follow that?

Incidentally, the process of solving this system will tell you if you used the right initial guess for your partial fraction expansion. You should get one unique solution. Otherwise, you'll need to change your initial guess. - July 12th 2010, 11:10 AMMechEng
I think I'm missing something here, where did the values on the LHS come from?

Ah... I think I see what happened here...

Since there are no x^3's on the LHS, that eqn. is equal to zero. Since the x^2 on the LHS divided by the x^2 on the RHS equals 1, we have that eqn. equal to 1.

Am I following along correctly here? - July 12th 2010, 11:38 AMAckbeet
What we're really doing here is reading off coefficients. If I were to give you the expression , answer the following questions:

1. What is the coefficient of the term?

2. What is the coefficient of the (constant) term?

3. What is the coefficient of the term?

4. What is the coefficient of the term? - July 12th 2010, 11:41 AMMechEng
1. 9

2. -7

3. 0

4. 0

Are we examing only the LHS or RHS (in your previous post) in this case?

Thank you very much for your patience with me. - July 12th 2010, 11:44 AMAckbeet
Those answers are correct.

Now, with the equations I gave you in post #6, we're looking at both sides, and equating the two. The coefficient of on the LHS must equal the coefficient of on the RHS. The coefficient of on the LHS must equal the coefficient of on the RHS. And so on. - July 12th 2010, 11:54 AMMechEng
Ah...

I have no idea why that wasn't clear from the beginning...

Thank you. - July 12th 2010, 11:57 AMAckbeet
Hehe. I love the light-bulb moments! That's what tutors live for.

So, what do you get for your constants? - July 12th 2010, 12:10 PMMechEng
I get:

Or is that better expressed as?:

- July 12th 2010, 12:12 PMAckbeet
I'd probably pick the first expression. Query: how can you double-check your answer?

- July 12th 2010, 12:45 PMMechEng
The answer can easily be checked by adding up the constants.